Respuesta :
(a) This follows from the definition for the partial derivative, with the help of some limit properties and a well-known limit.
• Recall that for [tex]f:\mathbb R^2\to\mathbb R[/tex], we have the partial derivative with respect to [tex]x[/tex] defined as
[tex]\displaystyle \frac{\partial f}{\partial x} = \lim_{h\to0}\frac{f(x+h,y) - f(x,y)}h[/tex]
The derivative at (0, 0) is then
[tex]\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(0+h,0) - f(0,0)}h[/tex]
• By definition of [tex]f[/tex], [tex]f(0,0)=0[/tex], so
[tex]\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(h,0)}h = \lim_{h\to0}\frac{\tan^2(g(h,0))}{h\cdot g(h,0)}[/tex]
• Expanding the tangent in terms of sine and cosine gives
[tex]\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{h\cdot g(h,0) \cdot \cos^2(g(h,0))}[/tex]
• Introduce a factor of [tex]g(h,0)[/tex] in the numerator, then distribute the limit over the resulting product as
[tex]\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{g(h,0)^2} \cdot \lim_{h\to0}\frac1{\cos^2(g(h,0))} \cdot \lim_{h\to0}\frac{g(h,0)}h[/tex]
• The first limit is 1; recall that for [tex]a\neq0[/tex], we have
[tex]\displaystyle\lim_{x\to0}\frac{\sin(ax)}{ax}=1[/tex]
The second limit is also 1, which should be obvious.
• In the remaining limit, we end up with
[tex]\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)}h = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h[/tex]
and this is exactly the partial derivative of [tex]g[/tex] with respect to [tex]x[/tex].
[tex]\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h = \frac{\partial g}{\partial x}(0,0)[/tex]
For the same reasons shown above,
[tex]\displaystyle \frac{\partial f}{\partial y}(0,0) = \frac{\partial g}{\partial y}(0,0)[/tex]
(b) To show that [tex]f[/tex] is differentiable at (0, 0), we first need to show that [tex]f[/tex] is continuous.
• By definition of continuity, we need to show that
[tex]\left|f(x,y)-f(0,0)\right|[/tex]
is very small, and that as we move the point [tex](x,y)[/tex] closer to the origin, [tex]f(x,y)[/tex] converges to [tex]f(0,0)[/tex].
We have
[tex]\left|f(x,y)-f(0,0)\right| = \left|\dfrac{\tan^2(g(x,y))}{g(x,y)}\right| \\\\ = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)^2}\cdot\dfrac{g(x,y)}{\cos^2(g(x,y))}\right| \\\\ = \left|\dfrac{\sin(g(x,y))}{g(x,y)}\right|^2 \cdot \dfrac{|g(x,y)|}{\cos^2(x,y)}[/tex]
The first expression in the product is bounded above by 1, since [tex]|\sin(x)|\le|x|[/tex] for all [tex]x[/tex]. Then as [tex](x,y)[/tex] approaches the origin,
[tex]\displaystyle\lim_{(x,y)\to(0,0)}\frac{|g(x,y)|}{\cos^2(x,y)} = 0[/tex]
So, [tex]f[/tex] is continuous at the origin.
• Now that we have continuity established, we need to show that the derivative exists at (0, 0), which amounts to showing that the rate at which [tex]f(x,y)[/tex] changes as we move the point [tex](x,y)[/tex] closer to the origin, given by
[tex]\left|\dfrac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}\right|,[/tex]
approaches 0.
Just like before,
[tex]\left|\dfrac{\tan^2(g(x,y))}{g(x,y)\sqrt{x^2+y^2}}\right| = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)}\right|^2 \cdot \left|\dfrac{g(x,y)}{\cos^2(g(x,y))\sqrt{x^2+y^2}}\right| \\\\ \le \dfrac{|g(x,y)|}{\cos^2(g(x,y))\sqrt{x^2+y^2}}[/tex]
and this converges to [tex]g(0,0)=0[/tex], since differentiability of [tex]g[/tex] means
[tex]\displaystyle \lim_{(x,y)\to(0,0)}\frac{g(x,y)-g(0,0)}{\sqrt{x^2+y^2}}=0[/tex]
So, [tex]f[/tex] is differentiable at (0, 0).
